I'm reading the book "Abstract Algebra with Applications" by Audrey Terras and one of the exercises (2.4.11) has us find the centers of $D_5$ and $D_6$ (the dihedral groups).
It states then, as a hint: The center of any group is contained in the center of any of its subgroups.
Either I can't figure out exactly what this hint means, or I just don't see how it could be true.
Let $E$ be the identity symmetry element, $R$ be a rotation of 90 degrees, and $F$ a flip or mirroring about some axis. The center of $D_4$ (which is $\{E, R^2\}$) is certainly not contained in "any" of its subgroups, for example the subgroup $\{E, F\}$.
Am I misunderstanding something here?
Best Answer
As has been noted, as stated the hint is very, very wrong. For example, $S_3\times C_2$ has center $\{e\}\times C_2$; the subgroup $S_3\times \{e\}$ has trivial center, so the center of the whole group is not contained in the center of this subgroup.
There are two possible corrections that might make it true.
As suggested by @ancientmathematician, they could have meant that the intersection of the center with any subgroup is contained in the center of the subgroup: $Z(G)\cap H\subseteq Z(H)$ for every subgroup $H$. This is certainly true. In the example above, it would tell you that the center does not contain any nontrivial elements of $S_3\times\{e\}$.
Alternatively, they could have been trying to say that the center of the group is contained in the centralizer of any subgroup. If $H$ is a subgroup of $G$, its centralizer is $$C_G(H) = \{g\in G\mid gh=hg\text{ for all }h\in H\}.$$ And it is easy to verify that $Z(G)\subseteq C_G(H)$ for every subgroup $H$.
(I will note that it is also false that the center of a subgroup is contained in the center of the group; in $S_3$, every proper subgroup equals its center, but the whole group has trivial center.)